Solution for 36217
Wanjie Wang
Teacher: Jiashun Jin
September 15, 2009
Ten points for each problem.
If there is any question, please contact me at [email protected]
1.Proof:
P
(
A
∩
B

B
) =
P
((
A
∩
B
)
∩
B
)
P
(
B
)
=
P
(
A
∩
B
)
P
(
B
)
=
P
(
A

B
)
2. The sample space of two dice rolled is Ω =
{
(
a,b
)

1
≤
a
≤
6
,
1
≤
b
≤
6
.
}
.
They have equal probability. So, the answers are:
P
(sum = 2)
= 1
/
36
P
(sum = 3)
= 2
/
36
P
(sum = 4)
= 3
/
36
P
(sum = 5)
= 4
/
36
P
(sum = 6)
= 5
/
36
P
(sum = 7)
= 6
/
36
P
(sum = 8)
= 5
/
36
P
(sum = 9)
= 4
/
36
P
(sum = 10) = 3
/
36
P
(sum = 11) = 2
/
36
P
(sum = 12) = 1
/
36
3.
P
(at least one dice is6) =
P
(the ﬁrst dice is six, the second dice is any number)
+
P
(the second dice is six, the ﬁrst dice is any number)

P(both dices are six)
= 1
/
6 + 1
/
6

1
/
6
*
1
/
6
= 11
/
36
4. According to the results above, the conditional probabilities are:
P
(
a
= 6

sum
= 7) =
P
(
a
sum
=7)
P
(
sum
=7)
=
P
((
a,b
)=(6
,
1))
P
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '09
 Jin
 Conditional Probability, Probability theory, white ball

Click to edit the document details